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Each of seven consecutive integers, all greater than positive integer [#permalink]
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30 Apr 2015, 04:23
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38% (02:24) correct 62% (02:42) wrong based on 185 sessions
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Each of seven consecutive integers, all greater than positive integer [#permalink]
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01 May 2015, 01:41
#1 N = 3 seems like a good contender possible list 0 1 2 0 1 2 0  good N = 2 doesn't work, coz we will get at least 3 odds there N = 4 doesn't work either coz the sequence looks like this: even odd even odd even odd even odd even odd ... there will always be at least 3 odds in the sequence of 7. N = 5: even odd even odd even even odd even odd even even 4 0 1 2 3 4 0  works N = 7: even odd even odd even odd even even odd even odd even odd even even  can't make a sequence of 7 with the required conditions N higher than 7 or even values won't work either Insufficient
#2 This statement alone is clearly insufficient, coz list can be like this: 0 1 2 3 4 5 0 or 0 1 2 3 4 5 6 which would mean that N is either 6 or bigger than 6
#1 + #2 gives us N = 5  sufficient
C



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Re: Each of seven consecutive integers, all greater than positive integer [#permalink]
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01 May 2015, 13:01
C St 1 n=3  120120 satisfies, also n=5 401234 satisfies St 2 odd number only once appears in the list; there can be so many numbers with rem non repeating odd numbers, n=7 1234560 and so on St 1+ St 2 n=3 has 2 odd repeating numbers so ans = 5



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Re: Each of seven consecutive integers, all greater than positive integer [#permalink]
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04 May 2015, 03:32
Bunuel wrote: Each of seven consecutive integers, all greater than positive integer n, is divided by n. The seven resulting remainders, not necessarily distinct, are assembled into a list L. What is the value of n?
(1) List L contains two odd values and 5 even values.
(2) No odd integer appears in list L more than once.
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:First, just get the info straight. You’re looking for n, which is a positive integer. There are also 7 other integers. Those 7 integers are both larger than n and consecutive, but n may not be consecutive with the rest. STATEMENT (1) NOT SUFFICIENT: Try some actual numbers in order to understand how all of this works. Attachment:
bgmat.png [ 2.84 KiB  Viewed 2072 times ]
At this point, it’s useful to think about what patterns exist for remainders. First of all, any remainders have to be less than the divisor. If the divisor is 3, then the largest possible remainder is 2. Second of all, for a given n, the pattern of remainders will be consistent, but that pattern could start on a different number. For example, above, the n = 3 pattern started with 0. If list L were 7, 8, 9, 10, 11, 12, 13, though, then the remainders would be 1, 2, 0, 1, 2, 0, 1. The pattern is the same; the sequence just starts with 1 instead of 0. Okay. So for n = 4, the pattern would be 0, 1, 2, 3, 0, 1, 2. The number 3 was added to the set. Further, the pattern could start on any number, depending upon the exact numbers in List L. As n increases by 1, one new remainder gets added to the pattern. Start thinking about what would need to be true in order to get two odd values and five even values. Have any of the patterns above already returned this result? Yes! When n = 3 and the pattern starts on the number 0, then there are exactly two instances of the odd value 1. How come that worked for n = 3 but not for n = 4? Interesting! In n = 3, the pattern is even, odd, even, even, odd, even, even. There are more evens than odds, so it’s possible to have only 2 odds. The pattern for n = 4, though, is even, odd, even, odd, even, odd, even. This time, there’s no way to get only 2 odds while having 5 evens. Okay, so for statement 1 to be true, the pattern has to be “asymmetric,” with more evens than odds. Where else is that going to happen? Not for any n = even cases. (Not sure about that? Try out n = 6 and see.) The variable n, then has to be odd. You already saw that n = 3 works. What about n = 5? The remainder pattern would be 0, 1, 2, 3, 4, 0, 1, 2, 3, 4. If you start the pattern at 4 (so that you have two evens in a row), then you’d have 4, 0, 1, 2, 3, 4, 0. Bingo! Exactly two odds. You’re done with this statement (for now, anyway) because there are at least two possible values for n. Statement 1 is not sufficient. STATEMENT (2) NOT SUFFICIENT: Many values of n are possible with this statement. For instance, if n is any integer greater than 6, then seven consecutive integers may be found to produce the remainders 0, 1, 2, 3, 4, 5, and 6, in which no integer (regardless of odd/even) appears more than once. STATEMENTS (1) AND (2) SUFFICIENT: We know that n = 3 and n = 5 satisfy the first statement. Are other odd numbers possible? No, it turns out. If n = 7, then the remainder pattern is 0, 1, 2, 3, 4, 5, 6. That list contains three odd numbers, not just two. For any larger n, the remainder pattern will have to include at least three odd numbers because the remainder pattern does not repeat within any set of 7 numbers. You’re not going to be able to produce the odd, even, even pattern that allowed n = 3 and n = 5 work. (Not sure about that? Try n = 9 and n = 11 to see how it works.) So only 3 and 5 are possible values for n, according to statement 1. If n = 3, though, the only possible odd remainder is 1, so the two odd numbers in list L must both be equal to 1. Therefore, n = 3 cannot satisfy the second statement, leaving only n = 5. When n = 5, the two odd integers 1 and 3 can each appear exactly once. The list has two distinct odd values and 5 even values, so n does indeed equal 5. The two statements together are sufficient. The correct answer is C.
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Re: Each of seven consecutive integers, all greater than positive integer [#permalink]
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13 Oct 2015, 01:27
Bunuel wrote: Bunuel wrote: Each of seven consecutive integers, all greater than positive integer n, is divided by n. The seven resulting remainders, not necessarily distinct, are assembled into a list L. What is the value of n?
(1) List L contains two odd values and 5 even values.
(2) No odd integer appears in list L more than once.
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:First, just get the info straight. You’re looking for n, which is a positive integer. There are also 7 other integers. Those 7 integers are both larger than n and consecutive, but n may not be consecutive with the rest. STATEMENT (1) NOT SUFFICIENT: Try some actual numbers in order to understand how all of this works. Attachment: bgmat.png At this point, it’s useful to think about what patterns exist for remainders. First of all, any remainders have to be less than the divisor. If the divisor is 3, then the largest possible remainder is 2. Second of all, for a given n, the pattern of remainders will be consistent, but that pattern could start on a different number. For example, above, the n = 3 pattern started with 0. If list L were 7, 8, 9, 10, 11, 12, 13, though, then the remainders would be 1, 2, 0, 1, 2, 0, 1. The pattern is the same; the sequence just starts with 1 instead of 0. Okay. So for n = 4, the pattern would be 0, 1, 2, 3, 0, 1, 2. The number 3 was added to the set. Further, the pattern could start on any number, depending upon the exact numbers in List L. As n increases by 1, one new remainder gets added to the pattern. Start thinking about what would need to be true in order to get two odd values and five even values. Have any of the patterns above already returned this result? Yes! When n = 3 and the pattern starts on the number 0, then there are exactly two instances of the odd value 1. How come that worked for n = 3 but not for n = 4? Interesting! In n = 3, the pattern is even, odd, even, even, odd, even, even. There are more evens than odds, so it’s possible to have only 2 odds. The pattern for n = 4, though, is even, odd, even, odd, even, odd, even. This time, there’s no way to get only 2 odds while having 5 evens. Okay, so for statement 1 to be true, the pattern has to be “asymmetric,” with more evens than odds. Where else is that going to happen? Not for any n = even cases. (Not sure about that? Try out n = 6 and see.) The variable n, then has to be odd. You already saw that n = 3 works. What about n = 5? The remainder pattern would be 0, 1, 2, 3, 4, 0, 1, 2, 3, 4. If you start the pattern at 4 (so that you have two evens in a row), then you’d have 4, 0, 1, 2, 3, 4, 0. Bingo! Exactly two odds. You’re done with this statement (for now, anyway) because there are at least two possible values for n. Statement 1 is not sufficient. STATEMENT (2) NOT SUFFICIENT: Many values of n are possible with this statement. For instance, if n is any integer greater than 6, then seven consecutive integers may be found to produce the remainders 0, 1, 2, 3, 4, 5, and 6, in which no integer (regardless of odd/even) appears more than once. STATEMENTS (1) AND (2) SUFFICIENT: We know that n = 3 and n = 5 satisfy the first statement. Are other odd numbers possible? No, it turns out. If n = 7, then the remainder pattern is 0, 1, 2, 3, 4, 5, 6. That list contains three odd numbers, not just two. For any larger n, the remainder pattern will have to include at least three odd numbers because the remainder pattern does not repeat within any set of 7 numbers. You’re not going to be able to produce the odd, even, even pattern that allowed n = 3 and n = 5 work. (Not sure about that? Try n = 9 and n = 11 to see how it works.) So only 3 and 5 are possible values for n, according to statement 1. If n = 3, though, the only possible odd remainder is 1, so the two odd numbers in list L must both be equal to 1. Therefore, n = 3 cannot satisfy the second statement, leaving only n = 5. When n = 5, the two odd integers 1 and 3 can each appear exactly once. The list has two distinct odd values and 5 even values, so n does indeed equal 5. The two statements together are sufficient. The correct answer is C. simple rules used to make a complex question



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Re: Each of seven consecutive integers, all greater than positive integer [#permalink]
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14 Oct 2015, 06:22
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Each of seven consecutive integers, all greater than positive integer n, is divided by n. The seven resulting remainders, not necessarily distinct, are assembled into a list L. What is the value of n? (1) List L contains two odd values and 5 even values. (2) No odd integer appears in list L more than once. There are 2 variables from this question (the first integer in the sequence and the number n), and we are given 2 equations from the 2 conditions. Therefore, there is high chance that the answer will be (C). Looking at the conditions together, the sequence becomes 4,0,1,2,3,4,0, and n=5. And if we give an example of 7 consecutive integers, we can have 9,10,11,12,13,14,15. The answer is therefore (C). For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Each of seven consecutive integers, all greater than positive integer [#permalink]
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04 Sep 2017, 03:41
MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Each of seven consecutive integers, all greater than positive integer n, is divided by n. The seven resulting remainders, not necessarily distinct, are assembled into a list L. What is the value of n?
(1) List L contains two odd values and 5 even values.
(2) No odd integer appears in list L more than once.
There are 2 variables from this question (the first integer in the sequence and the number n), and we are given 2 equations from the 2 conditions. Therefore, there is high chance that the answer will be (C). Looking at the conditions together, the sequence becomes 4,0,1,2,3,4,0, and n=5. And if we give an example of 7 consecutive integers, we can have 9,10,11,12,13,14,15. The answer is therefore (C).
For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. But even for your method you need to know the pattern of numbers which you cant know unless you try out. If i get questions like this in exam i would probably take a guess and mark some answer!!




Re: Each of seven consecutive integers, all greater than positive integer
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